WEBVTT
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When the leading coefficient of a quadratic expression is something other than one,
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there are two methods that we can use to factor the quadratic expression.
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The first method that I'm going to show you is called trial and error.
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Let's look at our first problem. Three r squared plus sixteen r plus five.
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This method of trial and error is the most efficient method when your values of a and c
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don't have many factors. So we're first going to begin by listing factors of five. Factors of five
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are plus and minus one and plus and minus five. Next, we will look at factors of a, or factors of three.
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I know that factors of three are plus and minus one and plus and minus three.
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Since we don't have a long list of factors for each of these numbers,
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trial and error is the method that will be most efficient. With trial and error, we are going to
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make our linear factors, and what we're looking for here is knowing that when we multiply these
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two numbers together we need a positive five. And so because we need a positive five,
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these numbers could either be negative and negative, or they could be positive and positive.
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But now we need to look more closely at our value of b. Since our value of b is a positive sixteen, the only
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way for us to get a positive sixteen sum and a product of positive five is if both of these numbers are positive.
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Now I know the only way for us to multiply these two together and
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get a positive five is if one of those numbers is five and the other is one. Now we need to figure out what these need to be
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so that when we multiply them together we will get a positive three r squared.
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Since it's r squared, I know that I will have to have an r in each of my factors, and the only way for us to get three r squared
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is if one of these numbers is three and the other is just one. So let's check to see if this is
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the correct factored form for this trinomial. Three r times r does give me three r squared.
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But notice three r plus five r does not add to give me sixteen r. Therefore, I need to know
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or I notice that this is not the correct factored form. So let's try another one.
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Again, I know that we will have to have a positive five and a positive one.
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But since three in this term and one in that term did not work, let's
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try one r there and three r here. To check, r times three r, three r squared.
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This is r plus fifteen r which does give me my sixteen r, and positive five times positive one does give me positive five.
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Therefore, this is the correct factored form of this trinomial. Now let's take a look at our second one.
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Again, notice your value of a is some number other than one. Again, I'm looking at factors of c as well as factors of a.
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For my factors of negative one, I know that that's just plus or minus one.
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And in terms of factors of four, I know that those are plus and minus one
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and plus and minus four as well as plus or minus two. So let's start trying to see which
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way we can factor this trinomial. Now since when multiplying these two numbers together I need to get a negative one,
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that tells me the only way to multiply two numbers together to get
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negative one is if one of them is positive one and one of them is negative one.
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Now let's try to work with these. I need four x squared, positive four x squared. One way
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to get positive four x squared is a positive two x and a positive two x.
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Now let's check to see if that is the correct form. Four x squared. Minus two x plus two x,
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that makes zero, not positive three. Therefore, this is not the correct factored form.
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Now we can try another set of factors of four. Again, this will have to be positive one and
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negative one. But now we can try a positive four x and a positive one x. Let's see if that's it. Four x squared.
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Minus four x plus x gives me a negative three x. That's not it. I need positive three x.
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So the only other possible way this could factor, again keeping the plus one and
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the minus one, this time let's try the one x here and the four x here.
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Four x squared. Minus x plus four x, there's my positive three x. And one times negative one is
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negative one. That means this, and not the previous two, is factored form of this trinomial.
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The second method that we can use in factoring a trinomial with a coefficient other than one
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is sometimes referred to as splitting the linear term. Let's take a look at our first example. Seven
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m squared minus nineteen m minus six. To begin this process, we are going to multiply a times c.
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Seven times negative six gives me negative forty-two. So a times c is equal to negative forty-two.
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It's also important to know the value of b which in this case is negative nineteen. Now we're looking for a pair of numbers
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whose product is negative forty-two and whose sum is negative nineteen. Let's begin by listing
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the factors of negative forty-two. Those are plus or minus one, plus or minus forty-two,
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plus and minus two, plus and minus twenty-one, plus and minus three,
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plus and minus fourteen, and our last set of factors plus or minus six and plus or minus seven.
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Now of this list of factors, we're most interested in the pair that adds to give
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us negative nineteen. Hopefully you can recognize that that is positive two and negative twenty-one.
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Now what we're going to do is we're going to split our linear term using
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these two numbers, so instead of negative nineteen m, I'm going to split my linear term into two m minus twenty-one m.
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Please recognize that rewriting it this way is not changing the value of the expression.
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It's simply rewriting it in a different form. I'm going to keep c, and I'm going to keep a.
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Now notice we have four terms, and we can factor four terms using the grouping method.
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I recognize that my first two terms do have a common factor, and I also recognize that
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my second two terms have a common factor. So I'm going to include
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the negative with the twenty- one, and I'm going to put addition between them. The common
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factor of my first group is m. Factoring out an m will leave me with seven m plus two.
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In my second group, I notice that there is a common factor of three. However,
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if I'm factoring out a positive three, then that will give me different signs than
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a positive seven m and a positive two. I'm going to factor out a negative three.
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Factoring a negative three out of this group will give me a positive seven m plus two. Now if you recall
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from the grouping method, both of these terms have a common factor of seven m plus two,
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and once I factor out seven m plus two from each of the terms, I am left with m minus three.
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And this is factored form of this trinomial. Let's look at our second example. Eight
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w squared minus eighteen w plus nine. When I multiply a times c, I get seventy-two.
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I notice that b is negative eighteen. Again, we're looking for factors of seventy-two
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whose sum is negative eighteen. So let's list the factors of seventy-two. They are plus or minus one
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and plus or minus seventy-two, plus or minus two and plus or minus thirty-six,
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plus or minus three and plus or minus twenty-four, plus or minus four, plus or minus eighteen,
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plus or minus six, plus or minus twelve, and finally plus and minus eight and plus
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and minus nine. Of these pairs of factors, the pair whose sum is also negative eighteen
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is negative twelve and negative six. So now I'm going to split my linear term into
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negative twelve w minus six w. I'm going to keep my first term, and I will keep my last.
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Again, recognizing that we now have four terms, I can now factor by grouping.
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I see that my first two terms do share a common factor as do my second two.
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The greatest common factor of the first group is four w, and when I factor out four w, I'm left with two w minus three.
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The greatest common factor of my second group happens to be three. However, if I factor out a positive three,
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then that will leave me with a negative two w and a positive three. I need a positive two w and a
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negative three. So I'm going to factor out a negative three. And when I factor out a negative three, I do indeed end up
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with two w minus three. Recognizing that both of these terms share a common factor of two w
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minus three, I can then factor out the two w minus three, and that makes my other linear factor
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four w minus three. And this is factored form of this trinomial. We're going to continue our practice of
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splitting the linear term with this next example. Let's factor twelve x cubed plus ten x squared plus two x.
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Notice that we do have a value of a that's other than one, but I also want you to notice
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that looking at all three of your terms you have a greatest common factor, and we should always
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begin the factoring process looking to see if we can factor out a greatest common factor.
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Looking at all three terms, I recognize that my greatest common factor is two x.
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If I factor out a two x from each of the terms, I'm going to be left with
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six x squared plus five x plus one. Now we're going to continue to factor this
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trinomial, and we're going to focus on the trinomial inside of the parentheses.
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This trinomial also has a leading coefficient other than one. So now
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we can choose whether to use the trial and error method or splitting the
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linear term. With a number like six which has more factors than say two, three, or four, or five,
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trial and error may not be the most efficient method in factoring this trinomial.
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So let's go ahead and use the method of splitting the linear term.
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a times c is six, and b is five. So I'm looking for factors of six who also add to give me five.
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Factors of six: plus or minus one, plus or minus six, plus or minus two, and plus or minus three.
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Of these factors, the pair which also adds to give me five is positive two and positive three.
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Therefore, I am now going to split my linear term and rewrite it as two x plus three x. I'm going to
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keep the one. I'm also going to keep the six x squared. Now I know that I have this factor, this greatest common factor
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of two x here which in the end we will absolutely put as part of completely factored form. Right now
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though, we're going to continue factoring these four terms using the grouping method.
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My first two terms have a common factor, and some of you may think the last two don't.
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However, they do, and we'll get to that in just a second. The common factor
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between six x squared and two x is two x, and if I factor out a two x, I'm left
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with three x plus one. Now let's talk about the common factor between three x and one.
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That's one. And so factoring out a one from three x plus one will give me three x plus one.
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Again, I notice a common factor here of three x plus one, and when factoring that from each of the terms,
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I'm left with two x plus one. Let's not forget our original greatest common factor of two x.
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And this is factored form of this trinomial.